proof of the weak Nullstellensatz

Let K be an algebraically closed field, let n≥0, and let I be
an ideal in the polynomial ringK⁢[x1,…,xn]. Suppose I is
strictly smaller than K⁢[x1,…,xn]. Then I is contained in a
maximal idealM of K⁢[x1,…,xn] (note that we don’t have to
accept Zorn’s lemma to find such an M, since K⁢[x1,…,xn] is
Noetherian by Hilbert’s basis theorem), and the quotient ring

by taking ϕ to be the identity on K and sending tj to uj.
Let N be the kernel of this homomorphism. Then ϕ can be
extended to the localizationK⁢[t1,…,tm]N of
K⁢[t1,…,tm]. Since ci,di∉N for all i, the
x¯i are integral over this ring. Since K is algebraically
closed, the extension theorem for ring homomorphisms implies that
ϕ can be extended to a homomorphism

ϕ:(K⁢[t1,…,tm]N)⁢[x¯1,…,x¯n]=L⟶K.

Because L is an extension field of K and ϕ is the identity on
K, we see that ϕ is actually an isomorphism, that m=0, and
that N is the zero ideal of K. Now let a1=ϕ⁢(x¯1),…,an=ϕ⁢(x¯n). Then for all polynomials f in the
ideal I we started with, the fact that f∈M implies